Problem: Find $ \lim_{x\to 1}f(x)$ for $f(x)=\sqrt{51-2x}$.
Solution: $f$ is a square-root function. Square-root functions are continuous across their entire domain, and their domain is all real $x$ -values for which the expression within the square-root is non-negative. In other words, for any square-root function $q$ and any input $c$ in the domain of $q$ (except for its endpoint), we know that this equality holds: $\lim_{x\to c}q(x)=q(c)$ [What happens at the endpoint?] The input $x=1$ is within the domain of $f$. Therefore, in order to find $ \lim_{x\to 1}f(x)$, we can simply evaluate $f$ at $x=1$. $\begin{aligned} &\phantom{=}f(x) \\\\ &=\sqrt{51-2x} \\\\ &=\sqrt{51-2(1)} \gray{\text{Substitute }x=1} \\\\ &=\sqrt{49} \\\\ &=7 \end{aligned}$ In conclusion, $ \lim_{x\to 1}f(x)=7$.